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Math Help - help! trig identity

  1. #1
    Member sinewave85's Avatar
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    help! trig identity

    I usually don't have any problem with these, but this one has me stuck.

    1 - sinA = (sin(A/2) - cos(A/2))^(2)
    subbing in the half-angle identities
    1 - sinA = (((1 - cosA) / 2)^(1/2) - ((1 + cosA) / 2)^(1/2))^(2)
    adding the complex fractions to simplify
    1 - sinA = (((1 - cosA)^(1/2) - (1 + cosA)^(1/2)) / 2)^(2)
    squaring (FOIL)
    1 - sinA = ((1 - cosA) - (1 - cosA) - (1 - cosA) + (1 - cosA))/2
    simplify
    1 - sinA = 2cosA/2
    not true
    1 - sinA = cosA

    It is tempting to say cos^(2)A = 1 - sin^(2)A, so cosA = 1 - sinA, but last time I checked, it does not work like that. CosA = (1 - sin^(2)A)^(1/2), and since that is a false square -- (1 - sinA)(1 + sinA) -- there is no real way to take the root. Tell me where I am wrong.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by sinewave85 View Post
    I usually don't have any problem with these, but this one has me stuck.


    1 - sinA = (sin(A/2) - cos(A/2))^(2)

    Expand the square:

    1 - sin(A) = sin^2(A/2) -2 sin(A/2)cos(A/2) + cos^2(A/2)

    ............. = 1 - 2sin(A/2)cos(A/2),

    so you need only show:

    sin(A) = 2sin(A/2)cos(A/2),

    which I hope looks familiar.

    RonL
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  3. #3
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    Hello, sinewave85!

    The Captain's method is the way to go, but your approach will work, too.


    1 - \sin A \:= \:\left[ \sin\left(\frac{A}{2}\right) - \cos\left(\frac{A}{2}\right) \right]^2

    Using half-angle identities, the right side becomes: . \left[\sqrt{\frac{1-\cos A}{2}} - \sqrt{\frac{1+\cos A}{2}}\right]^2

    Square: . \frac{1-\cos A}{2} \:- \:2\cdot\sqrt{\frac{1-\cos A}{2}}\cdot\sqrt{\frac{1-\cos A}{2}} \:+ \:\frac{1 + \cos A}{2}

    We have: . \left(\frac{1-\cos A}{2} + \frac{1+\cos A}{2}\right) \:- \:2\!\cdot\!\sqrt{\frac{1-\cos^2A}{4}}

    . . . . . . =\;1 \:- \:2\!\cdot\!\sqrt{\frac{\sin^2A}{4}}

    . . . . . . = \;1 \:- \:2\!\cdot\!\frac{\sin A}{2}

    . . . . . .  \;=\;1 - \sin A . . . ta-DAA!

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  4. #4
    Member sinewave85's Avatar
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    Thanks for the help!
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